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We are asked to find the minimum area of a square given that the area is expressed as A = 9x^2 + 12x +4.
We know that area cannot be negative, therefore we can make the following cases:
9x^2 + 12x + 4 > 0 or 9x^2 + 12x + 4 = 0.
Examining the two cases, we realize the minimum value would be 0 for the area of the square.
Therefore, the minimum value of the square is 0.
The area of the square is given as A = 9x^2 + 12x + 4
To minimize the area we solve the first derivative of A for x.
dA/dx = 0
=> 18x + 12 = 0
=> x = -12/18
When x = -12/18, A = 0
This proves what is evident from the fact that A represents the area, it cannot be negative. The minimum value that the area can take is 0.
The minimum area of the square is 0.
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